 Alright, so we are discussing the importance of local rings okay and what we saw in the last lecture was that we have seen two theorems which tell us the importance of local rings. So the first one was that a morphism is an isomorphism if and only if it is a homeomorphism and induces isomorphism at the level of local rings so that is a characterization of isomorphism in terms of local rings okay. Then the other thing that we saw was we saw that a rational function okay an element of the function field of our IT namely a function which is which may be defined only on a proper open subset that can be I mean you can ensure that it is actually a global regular function if and only if it occurs in the it occurs in every local ring considered as a subring of the function field okay. So in this regard what I wanted to tell you is that you know there is this algebraic fact which is simple algebraic fact and this is what we proved last time you see if A is an integral domain A is an integral domain and then of course A is contained in its localization at any maximal ideal okay. So max spec A is the set of all maximal ideals of A okay and of course every integral domain is contained in its localization at a maximal ideal this map A to its localization is just the localization map okay and it is injective and that is why we consider A as a subring of the localization of A at M and further this is contained in the quotient field of A which is the same as the quotient field of A of the localization of A at M okay. So this is so what this tell I mean this always happens alright and what this tells you is that A is contained in therefore the intersection of all the localizations at various maximal ideals inside the quotient field okay but the fact is that this is actually inequality if you go through the proof of the statement that we made last lecture that a rational function is a global regular function if and only if it occurs in every local ring okay then actually we have actually if you see that the commutative algebra that we did there the proof actually proves this that if A is an integral domain then A is exactly the intersection of all its localizations at various maximal ideals the intersection being made sense of in the quotient field of A. So this is a completely you know algebraic commutative algebraic fact but it has a geometric meaning geometric meaning is that if you take for A the if you take A to be the affine coordinate ring of an affine variety okay then its then this quotient field of A is nothing but the function field of the variety and you are saying an element of the function field which is in every local ring has to be in has to be a global regular function okay because the AM's will then be the local rings at the points AM for example will be the local ring of the variety with affine variety with coordinate ring A at the point corresponding to the maximal ideal M okay and at the fact that the intersection of all these local rings is A is a nice statement that when you translate it geometrically it means that a rational function which is there in every local ring is actually a global regular function okay so basically I just wanted to point out that this is an algebraic fact but it has a geometric meaning okay the geometric meaning is that a rational function which is in every local ring is actually defined everywhere it is a global regular function right okay so now what I stated at the end of the previous lecture was a fact about birationality so let me let me again recall that statement so the so this is yet another nice result that tells you about the importance of local rings so the result is that if you have two varieties which may not be related at all but you know suppose I know the local ring of at of one variety at one point is isomorphic as k algebra to the local ring of another variety at another point then the deep result is that these two varieties are isomorphic on open subsets containing the respective points which we call as birational okay so just isomorphism of local rings produces an isomorphism over a large open set because you know open sets are all large in the Zariski topology so this also tells you that the local ring is not so local it also contains lot of global information okay it contains information on large open set and that is in a way inevitable because in the Zariski topology all open sets are huge because they are all dense any open set is dense okay for a variety so of course any non-empty open set so so let me write this down so theorem let X and Y be varieties with X small X a point of capital X small Y point of capital Y and an isomorphism of k algebras from the local ring of X capital X at small X to the local ring capital Y at small Y so this is maybe I can call this as I do not know what I called it in the last lecture so let me call it as phi okay I do not know what what did I call it I called it psi okay so let me call it psi let me use the same notation so I have these two varieties and I have points where the local rings are isomorphic by a k algebra isomorphism psi okay then there exists open subsets U of X V of Y with X and U Y in V such that and an isomorphism of varieties phi from U to V that takes so that X goes to phi of X is Y okay X goes to Y and which induces via phi's phi hash so I will get a map from the local ring of V at Y to the local ring of U at X and this is this map is phi hash sub Y okay and mind you the local ring of V at Y is the same as local ring of Y at capital Y at small Y because the local ring does not change if you go to a an open set and similarly this is the local ring of capital X at small X okay so phi is an isomorphism of varieties therefore this will also be an isomorphism and the fact is that this isomorphism is this isomorphism is none other than psi inverse okay the k algebra isomorphism given k algebra isomorphism phi hash Y equal to psi inverse okay so in other words any isomorphism so what I am saying is you see if a local ring of one variety at one point is isomorphic to a local ring of another variety at another point then that isomorphism is actually induced on from an isomorphism on an on open subsets of an open neighbourhood of U of the point small X and an open neighbourhood V of the point small Y they are isomorphic okay so the isomorphism of local rings comes from an isomorphism of huge open sets which contain those points so in other words see isomorphism at the local ring actually if you think of local ring as concentrating attention at a point what you are saying is that isomorphism of these local rings has I mean it can actually be spread to an isomorphism on open sets okay so that is the power of so you know so and you know when we have two varieties which have opens which are isomorphic on open subsets okay then we call those varieties as birational and what we are saying here is that of course if two varieties are birational then there are going to be local rings here which are going to be there are going to be isomorphic local rings okay wherever the isomorphism is defined but conversely if you want two varieties to be birational all you require is one isomorphism between local rings okay so that is the power of local rings so let us try to prove this fact so the proof also involves just commutative algebra and several facts that we have already seen so here is the proof but so the first so what is given to me is that I am given x I am given y and I am given a point small x in capital X and I am given a point small y in capital Y okay and what I am given is at the level of local rings I am given an isomorphism so well if I draw the diagram corresponding to this here I will get well x corresponds to O x and y corresponds to O y so you see here I am writing the well if you want this is the this is the geometric side and this is algebraic side okay so then I have corresponding to this point x I have the local ring O xx O capital X small x and here I have O capital Y small y and then I have this isomorphism which is given to me which is given to me as psi it is a k-algebra isomorphism okay this is what I have alright and what I am supposed to do is I am supposed to find an open set U here which contains x and an open set V here which contains y and an isomorphism between U and V which induces at the point x is isomorphism okay so my the first fact I will use is that any variety is covered by finitely many open subsets which are isomorphic to affine varieties okay so what I will do is that I will instuff x I will look at x1 inside x where x1 is an affine open x1 is an affine open which contains the point small x okay and similarly I can find an affine open y1 I can find an affine open y1 and affine open subset of y which is affine open means it is an open subset of y which is isomorphic to an affine variety and which contains the point small y okay so here I am using the fact that we have proved earlier that any variety can be covered by finitely many open sets each of which is isomorphic to an affine variety right and well what happens to what happens to this diagram on this side see I will have I will have this so what you must understand is that there is this inclusion of the regular functions are included in the local ring okay and in fact what will happen is that if you if you go to O or if you go to O x1 that will be that will sit inside that will sit in between okay because you know for any variety restriction of regular functions then is an injective homomorphism okay because if I have a regular function on a bigger open set I can restrict the regular function to a smaller open set and this restriction is injective because if two regular functions agree on a smaller open set then they agree on the bigger open set okay. So this will and mind you this O x1 will be well it is it is going to be A x1 I can write O x1 equal to A x1 because x1 is affine because x1 is affine and well I have the same kind of situation here I have O y1 this is sitting inside this and this is sub ring of this and this is the same as A y1 okay and of course and what about the local rings see the local ring of x capital X at small x will be the same as the local ring of capital X1 at small x because the local ring will not change if you go to enough affine open set and then in this identification with A x1 this local ring is going to correspond to some A so you know let me do something let me call this as A yeah I must have had little bit more space to write things so let me move part of this diagram to the left side so that is easier so here is y and here is y1 and here is small y okay so this A x1 let me call it as A let me call A y1 as B okay and see what you must understand is that this x1 is an affine variety therefore you know this x1 can be embedded into some affine space into some A n okay of course we are working over small k which is an algebraically close field always and similarly this y1 will sit inside some A m okay so when I write like this what I mean is that x1 is isomorphic to an open I mean a close irreducible close subset of A n and y1 is isomorphic to an irreducible close subset of A m okay and therefore what you must understand is that the points the points of x1 will correspond to maximal ideals in A and the points of y1 will correspond to maximal ideals in B and the point small x will correspond to a certain maximal ideal and the local ring will be just A localized at that maximal ideal okay and where the maximal ideal m corresponds to the point small x okay in this in this identification of capital x capital x1 with max spec A after all max spec A set of all maximal ideals in A by the Nulst-Stellen-Satz is identified with the points of x1 because A is A of x1 alright and similarly the so x corresponds to the point small x in capital x1 corresponds to a maximal ideal m of A and the local ring is then just the localization of A at m okay similarly y y corresponds to a maximal ideal neta which is which corresponds which comes again by this correspondence due to the Nulst-ellen-Satz I have max spec of B the maximal ideals of B correspond to points of y1 okay and the point small y of y1 corresponds to this maximal ideal neta and therefore the local ring here will be the same as local ring of y1 at y and that can be identified with B neta okay. So we have already seen this the local ring of an affine variety at a point is just given by taking the localization of the affine coordinate ring maybe the ring of regular functions at the maximal ideal that corresponds to that point right. So basically you know at the end of all this my diagram and you know what I am going to do is I am going to find an open subset of x1 and I am going to find an open subset of y1 and I am going to find an isomorphism between them okay and you must realize that is good enough because an open subset of x1 is also an open subset of x and open subset of y1 is also an open subset of y and therefore you have also found an isomorphism between an open subset of capital X and an open subset of capital Y okay which contain the points x and y okay. So in other words you know to say it in simple words without loss of generality I could have assumed that x and y are already affine okay. So anyway so my diagram is now so I simplify my diagram so I have this A and I have this B and I have A sub n I have B sub n and I have this psi which is an isomorphism of local series okay this is the situation and well you know I am therefore looking at only the I am only looking at the affine case so here in this so on this part of the diagram you know so let me put a line here and look at this part of the diagram I have x1 I have y1 x1 corresponds to A and y1 corresponds to B right so I have this situation. Now what I want you to understand is that the what about see if you look at quotient field of A this is the same as the quotient field of A m okay the quotient field of A is the same as the quotient field of A m and what is it is the function field of x1 so this is just kx1 okay and similarly the here the and of course A sits inside its localization and the localization sits inside its quotient field okay similarly B sits inside in its localization and that sits inside in the quotient field of B which is the same as the quotient field of B neta which is the same as the function field of y1 okay. So this is again this is the fact that the function field of an affine variety is just the quotient field of its ring of regular functions okay so x1 is an affine variety its ring of regular functions is A which is also the affine coordinate ring of x1 and if you take its quotient field what you will get is a function field of x1 okay and the function field of x1 is of course the equivalence classes of rational functions where rational functions are regular functions on open subsets okay. So now what you must understand is that whenever you have an isomorphism between two ring two integral domains then it will automatically induce an isomorphism of the quotient fields if two so this is very natural fact which comes out of the if you want the universal property of localization or the universal property of the quotient field which says that given a ring given an integral domain its quotient field is the smallest field which contains the given ring name so if the ring is embedded in another integral domain okay then if the ring is embedded in another field okay then its quotient field will also get embedded in that field so the quotient field is the smallest field in which an integral domain can sit okay so because of that property psi will induce an isomorphism here I will call this as I will also call this as psi so maybe I will so let me write this as induced by psi this is just induced by psi okay and this diagram commutes okay. So what you must remember is that it is even very easy to see what this map is okay what is an element here an element here is a fraction the numerator and element of a and the denominator and element of a which is outside m okay and that is how an element here looks like and what you will have to do is if you go to the quotient field the quotient field is simply quotients of elements of a with the denominator non-zero okay but you can take you can represent such a quotient as a quotient of two quotients here okay by taking the denominators as one and then you can define what this map is simply by using this map okay so it is pretty easy to see what this map is it is straight forward to define okay even if you do not want to worry about the universal property and things like that okay you can directly write out this map it is the most natural map that you can think of okay. So this isomorphism induces this isomorphism okay and now what we are going to do is now we are going to make use of the fact that these two are affine okay so you see a is well a is what a is the affine coordinate ring of x1 it is a ring of regular functions on x1 and that is the polynomials in a and the polynomial functions on a and restricted to x1 okay so a is just k of x1 etc xn mod low the ideal of x1 okay this is the affine coordinate ring of x1 right and similarly b is b is the affine coordinate ring of y1 y1 is embedded in am okay so b which is the affine coordinate ring of y1 is just the polynomials in am okay polynomials in m variables restricted to y1 and that is given by k of y1 etc okay so there is a conflict of notation somehow so let me okay so I have conflict of notation my x1 is also a variety and my x1 is also a variable so it is a pity so let me do the following thing. Let me change this to something which is more decent so let me use s1 etc up to sn where si are the variables coordinates and here let me use t1 etc tn divided by ideal of y1 okay you must remember that this ideal of x1 is the ideal inside this okay it is ideal in am and this is the ideal of y1 inside this which is in am okay this am here is taken with coordinates s1 through sn this am is taken with coordinates t1 through tm so this is tm okay so this so here the coordinates are s1, sn and here the coordinates are t1, tm okay so that is the situation and now you see now what am going to do is am going to look at so you know this a is generated by the images of the si's okay therefore am going to look at what happens to the effect of si on the si's and am going to also look at what happens to the effect of si inverse on the ti's okay so the situation is like this I calculate so let me write here calculate si of let us look at si of si bar divided by 1 okay so what you must understand is si is an element in this polynomial ring okay it is a variable in the quotient you get si bar okay and if you go to the localization you write it as si bar by 1 so the natural map from ring to its localization takes any element of the ring to the to that element divided by 1 okay this is the standard map into any localization right so any element here alpha will go to alpha by 1 so if I take the element si bar it will go to si bar by 1 so am looking at it here alright but then you know you can treat si bar by 1 also as si itself si bar itself because if you think of everything is happening in this quotient field I do not have to write this divided by 1 but anyway I will write it so that you remember that I am taking its image here and am applying si to it right so now what is this now the image of this is going to be something here alright so but something here is a fraction it consists of a numerator and denominator the numerator is an element of b the denominator is an element of b which is not in neta in the maximal ideal neta so what I am going to get is I am going to get si bar by gi bar okay where of course fi bar and gi bar are just polynomials in the tj's okay and gi bar is not in the maximal ideal neta that the maximal ideal neta of b that corresponds to the point small y okay so let me write that gi bar does not belong to neta okay so what you must understand is if gi bar is I mean gi bar is not in neta therefore so 1 by gi bar makes sense here 1 by gi bar makes sense here because it is after all an element which is inverted what is b sub neta b sub neta is take all the elements of b and then you divide you allow fractions where you invert things which are outside the maximal ideal neta and gi bar is not in neta so 1 by gi bar makes sense in b neta right but then this 1 by gi bar lives here okay so it has to come from something there because after all si is an isomorphism okay and so I so I can look at those elements here so you know let me write fi bar of tj by so you know I will just put fi bar of t by gi bar of t just to remind you that the fi's and gi's are you know polynomials in the t's okay and this and of course something here will be a polynomial in the s's okay so you let this and also you let psi inverse of 1 by gi bar of t to be something here and that something there is going to be an element in the in here okay and if I write this as well let me write this as ai bar ai ai of s bar by bi of s bar when I write s I mean the tuple of variables s1 through sn and when I write t I mean that m tuple of variables t1 through dm okay the things on this side are all polynomials in the s's and the things on this side are all polynomials in the t's and if you go to the level of quotient fields here the things on this side are essentially thought of quotients of polynomials in the s's and here the elements are quotients of polynomials in the t's okay so so now what I wanted to understand is that see whenever you have an isomorphism of local rings okay then you know local ring has only one maximal ideal okay so whenever you have an isomorphism from one local ring to another local ring then under isomorphism the image of a maximal ideal is again a maximal ideal therefore what it means is that the maximal ideal of this local ring the unique maximal ideal of this local ring has to be carried by this isomorphism onto the unique maximal ideal of that local ring okay and that means that the an isomorphism of local rings will map given isomorphism of the corresponding maximal ideals and the fact that 1 by gi bar is the fact that you know so I want to say that this when I write this expression okay I want to say that this fellow here is invertible okay because you see gi if you want 1 by gi bar of t is a unit here 1 by gi bar of t is a unit here because gi bar of gi bar gi bar is the gi bars are outside the maximal ideal therefore they all get inverted here so it is a unit and an isomorphism will carry a unit to unit therefore it means that ai bar by bi this ai bar by bi bar is a unit okay and since ai bar by bi bar is a unit it means that both ai bar and bi bar are units okay so the upshot of this is that both ai bar and bi bar do not belong to the maximal ideal m here okay okay where ai bar bi bar do not belong to the maximal ideal okay so this is a this is an observation this is an observation right this is an observation. So now let us interpret this let us interpret this little carefully now you see see somehow you know we have to go you know you want you have at the level of small x you have at small x and at small y at the local ring level you have an isomorphism you have to somehow lift it to an open set so I have to go closer to x1 and y1 so which means that I will have to come go away from the local ring and I have to come closer to this ring okay so you know if you look at what happens to the image of psi so the first thing is that you know if I take you see the psi takes each si bar into expression like this alright therefore if you take any but any element of a is a polynomial in the si bar okay because a is just polynomial in the si bar modulo is ideal so any element of a can be thought of as a polynomial in the si bar okay in the in other words we say that a is generated by the si bar okay therefore any element of a is a polynomial in the si bar and if you take its image under si okay then what I am going to get is a polynomial in these things with k coefficients okay and you can see that the denominators are given by the gis okay therefore you can see very clearly that the image of a under the image of the subring a under si lands inside this localization namely b localize that the product of all the gi bars okay so there is this localization b you invert all the gi bars and the product of all the gi bars you will have a further localization this will be a intermediate localization because you see all the gi bars are not in eta okay all the gi bars are not in eta and that means that the product of all the gi bars will also not be in eta okay because eta is a is a maximal ideal so it is prime therefore b localize that all the product of all the gi bars which is which by the way is the same as inverting each of the gi bars okay so this is b localize the product of all the gi bars and that is a that will sit inside a subring between b and b nature okay and the fact is that the if you follow a by this it actually lands inside this psi restricted to a okay psi restricted to a will be will land inside this because after all what is an element of a it will be a polynomial in the si bars and if you apply psi to each si bar I am going to get fi bar by gi bar so if I have a polynomial expression in the si bars I am going to get a polynomial expression in the fi bars by gi bars and that is some polynomial divided by some product of gi's okay some product of powers of gi's and therefore it will lie here alright so the moral of the story is that si restricted to a will give you an injective k algebra homo from this this right and you know the point I want to make is that what have we done what have we done geometrically see geometrically what you have done is you see you have gone to each gi bar is a polynomial is a regular function okay each gi bar is an element here but what are the elements here the elements here are just regular functions on y1 okay and what do you mean by inverting a gi bar inverting a gi bar amounts to going to the open set the basic open set given by the non vanishing of the gi bars okay when you invert a regular function okay you go to the when you when you invert a regular function the ring that you get is the affine coordinate ring of the basic open set defined by that regular function namely the locus where that regular function does not vanish so what you are getting here is the basic open set given by the product of the gi bar that is a this is the locus in y1 where none of the gi bars vanish okay and that locus contains the point small y okay that is because none of the gi bars vanish is at y and that is that is just the resatement of the fact that none of the gi bars is in the maximal ideal neta which is where neta is the maximal ideal corresponding to the point small y okay therefore what has happened is that you know we have gotten hold of an open set here alright and we have in fact a morphism of varieties from this open set to x1 okay we have a morphism like this and that is because you see uhhh psi restricted to a psi restricted to a is just uhhh from a to psi of a is an isomorphism because after all psi is psi is an isomorphism so the restriction is again an isomorphism and you see uhhh uhhh and psi of a is contained in uhhh uhhh b sub product of gi bar okay and mind you this is a uhhh uhhh this is just a of x1 okay and this is the thing here is a of d the basic open set defined by the product of gi bar okay and the moment you have a k algebra home morphism from the affine coordinate ring of one affine variety to the affine coordinate ring of another affine variety it corresponds to a morphism in the reverse direction because we have seen this uhhh that there is a bijective correspondence between the morphisms of two affine varieties and the k algebra home morphisms between the affine coordinate rings therefore the psi restricted to a you know that is a k algebra home morphism from a to psi a and therefore it is a k algebra home morphism from a to b uhhh localized product of gi bar therefore it induces a map like this uhhh which is a morphism of varieties and that is this map that is this map and so uhhh so this is uhhh uhhh well I do not know where I should give a give a name to this uhhh so let let me write this uhhh I will just write this as induced uhhh by psi restricted to a okay there is a map like this and now whatever I did uhhh with uhhh after all a and b uhhh psi is an isomorphism whatever I did with a now I can do it with b okay so what do I how do I do the same thing uhhh with b so uhhh again what you do is that you know you look now instead of looking at psi you look at psi inverse okay so again calculate uhhh let us write out psi inverse on uhhh on on on generators of b what are generators of b the ti bar the tj bar generate b okay so calculate the image of uhhh tj bar under uhhh psi inverse okay so tj bar is the image of tj in this quotient ring okay and it is thought of as the element tj bar by 1 here and you are applying psi inverse and you are going to get something here alright and that is going to be simply a quotient of polynomials uhhh in the s's with the denominator outside the maximal ideal okay so you are going to get this is equal to f i of uhhh uhhh fj of s divided uhhh bar by gj of s where uhhh gj uhhh bar uhhh is not in the maximal ideal m you are going to get this okay when I when I take when I take the image of uhhh tj bar here right it is going to be an element here alright and an element here is a fraction it is a quotient of 2 elements the numerator any element of a and the denominator an element of a which is outside m and of course elements of a are just polynomials in s okay red model of the ideal of x1 okay so I am going to get this and again if you calculate uhhh uhhh uhhh psi inverse of of uhhh so if you calculate psi of 1 by gj bar uhhh this is going to give me uhhh 1 by gj bar is find you a unit here okay because gj bar are uhhh polynomials are not in the maximal ideal m okay elements which are not in the maximal ideal become units in the localization at that maximal ideal and when you apply psi to that I am supposed to get a unit here okay and I I will I will write it as a i bar uhhh a j bar of t uhhh by uhhh bj bar of t with both a j bar and bj bar with both a j bar and bj bar not in uhhh the not in the maximal ideal meter okay because that is how units will look like right. So you are going to get something like this and if you repeat the same thing that you did uhhh for psi for psi inverse what you are going to get is a following uhhh so what this tells you is that sign the image of the subring b under psi inverse will go to the localization of a at the product of the gj's okay so you know I am going to have something like this uhhh so so I have psi inverse in this direction alright and b is a subring here when I take the image of b I am going to land in the localization of a at the product of all the gj's so I will uhhh so let me write that I have a product of the gj bar this is localization of a at the product of the gj bars and uhhh this sits inside this and this of course it is inside am because all the gj bars are outside uhhh uhhh all the gj bars are units in the uhhh in this local ring okay because they are outside the maximal ideal m all the gj bars are outside m okay therefore uhhh this localization sits inside this localization and uhhh what is going to happen is that if I apply psi inverse to b it is going to land inside this so I am I am going to get something like this I am going to get an arrow like this which is just psi inverse restricted to b okay because when I apply psi inverse to something here something here is just a polynomial in the t bars in the tj bars and a polynomial in the tj bars under if I if I apply psi inverse I will get a polynomial in the fj bars by gj bars okay. So it is going to be some polynomial in the s divided by some product of powers of gj's so it is going to land inside this okay and again uhhh uhhh so again you know just as in the earlier case you can write psi inverse restricted to b is going to give you map from b uhhh it is an isomorphism of b with psi b and uhhh psi b is contained in uhhh a localize at the product of gj bar okay and uhhh but b is just the affine coordinate ring of y1 and this is a uhhh localize at the product of gj bar is affine coordinate ring of the basic open set where gj bar do not vanish. So this is just a of d of product of uhhh gj bar okay so the moral of the story is that uhhh uhhh just as in this case uhhh uhhh you get a morphism from uhhh d product of g a bar to x1 you here also you will get a product a morphism from uhhh the basic open set depend by product of small gj bar uhhh and uhhh uhhh and into y1 okay. So the moral of the story is I will if I translate this psi inverse restricted to b from b to this I will get a I will I will get a map like this I will get I have d product of gj bar which is an open subset of x1 which contains the point x mind you the point x uhhh uhhh belongs to uhhh this open set that is because none of the gj bars belong to the maximal ideal corresponding to x okay. So x is in the locus of non vanishing of the gj bars and the non vanishing of the gj bars at x is uhhh uhhh geometric uhhh expression of the fact that the gj bars are not in the maximal ideal corresponding to x okay. So so so as a result I will get a morphism like this I will get a morphism like this and this morphism is uhhh this is this this morphism is induced by psi inverse restricted to b okay. So you know uhhh I am coming close to trying to find uhhh see basically I want an open subset of x1 and an open subset of y1 and I want an isomorphism between them. So uhhh what we have done is we have gotten open sets here and there and we have got uhhh morphisms into the ambient space we will have to make sure you have to make this open sets smaller uhhh to check that you get morphisms between them okay and that is where the big deal comes. So you know uhhh now what you do is you uhhh I have still I have still things uhhh to be inverted on the on the side of the T's I mean on the side of the S's namely the A i's and B i's and I still have things to be inverted on the side of the of the T's namely the small aj's and the small bj's okay. So what you do is you take you go to further localization okay you here you go to the further localization of A at product of gi g j bar and you also take the product of product over all capital A i bar capital B i bar this is a further localization okay which sits inside this ring alright. Similarly you go here to a further localization namely you take B this is already localize at product of gi bar but you still have these fellows small aj bar small bj bar which are outside neta. So I can take the product of small aj bar small bj bar okay and that is a further localization and that that is also sits inside this okay and what does it what does it uhhh uhhh amount to what is what it amounts to is here you are going to smaller open set namely you are going to open set given by product of gi bar into product of capital A i bar B i bar which is an which is open inside this and x is actually here okay and here you are going to a smaller open set namely the localize where uhhh the product of capital GI bar and product of small aj bar bj bar and y is here.